Area and perimeter of a square word problems are much useful to kids who would like to practice problems in mensuration.
(1) Find the area of the square having side length 24 cm.
(2) A square is of area 64 cm². Then find its side length.
(3) The square having side length 25 cm. Find the area in meter.
(4) Find the area of the square whose diagonal is measuring 4cm.
(5) The diagonals of two squares are in the ratio 2:5. Find the ratio of their area.
(6) Find the perimeter of the square having side length 24 cm
(7) Find the perimeter of the square having side length 15 cm
(8) A square is of area 64 cm². What is its perimeter?
(9) Find the perimeter of the square whose diagonal is measuring 4cm.
(10) The perimeter of two squares are 40 cm and 32 cm. Find the perimeter of third square whose area is equal to the difference of the area of two squares.
(11) The perimeter of five squares are 24 cm, 32 cm, 40 cm, 76 cm and 80 cm respectively. The perimeter of another square in area to the sum of the areas of those squares is :
Question 1 :
Find the area of the square having side length 24 cm.
Solution:
Area of the square = a² = 24²
= 24 x 24
= 576 cm²
Hence, area of square is 576 cm²
Let us see the solution of next question on "Area and perimeter of a square word problems".
Question 2 :
A square is of area 64 cm². Then find its side length.
Solution :
Area of the square = 64 cm²
a² = 64 cm²
a = √64
= √8 x 8 ==> 8 cm
Hence, side length of square is 8 cm
Let us see the solution of next question on "Area and perimeter of a square word problems".
Question 3 :
The square having side length 25 cm. Find the area in meter.
Solution :
Area of the square = a² = 25² ==> 25 x 25 ==> 625 cm²
100 cm = 1 m
= 625/100 ==> 6.25 m²
Let us see the solution of next question on "Area and perimeter of a square word problems".
Question 4 :
Find the area of the square whose diagonal is measuring 4 cm.
Solution :
The diagonal AC divides the square into two right triangles.Δ ACB and Δ ADC. In triangle ACB right angle is at B.
So the side which is opposite to right angle is called as hypotenuse.
By using Pythagorean theorem
AC² = AB² + BC²
4² = x² + x²
16 = 2x²
8 = x²
√8 = x
√2 x 2 x 2 = x
2√2 = x
Therefore, the length of each side is 2√2 cm
Area of the square = a²
= (2√2)²
= 2²(√2)²
= 4 (2)
= 8 cm²
Let us see the solution of next question on "Area and perimeter of a square word problems".
Question 5 :
The diagonals of two squares are in the ratio 2:5. Find the ratio of their area.
Solution :
Let the diagonals of two squares be 2x and 5x respectively.
Area of a square when diagonal is given = (1/2) x d²
Area of first square = (1/2) x (2x)²
= (1/2) x 4x² ==> 2x²
Area of second square = (1/2) x (5x)²
= (1/2) x 25x² ==> 25x² / 2
Ratio of their areas ==> 2x² : (25x² / 2)
= 4 : 25
Let us see the solution of next question on "Area and perimeter of a square word problems".
Question 6 :
Find the perimeter of the square having side length 24 cm
Solution :
Perimeter of a square = 4a
here a = 24 cm
= 4 ( 24 )
= 96 cm
Hence, perimeter of the square is 96 cm
Let us see the solution of next question on "Area and perimeter of a square word problems".
Question 7 :
Find the perimeter of the square having side length 15 cm
Solution :
Perimeter of a square = 4a
here a = 15 cm
= 4 ( 15 )
= 60 cm
Therefore perimeter of the square is 60 cm
Question 8:
A square is of area 64 cm². What is its perimeter?
Solution :
Area of a square = 64 cm²
a² = 64 cm²
a = √ 64
a = √8 x 8
a = 8 cm
Now we have to find the perimeter
Perimeter of the square = 4a
= 4 (8)
= 32 cm
Therefore perimeter of the square is 32 cm
Question 9 :
Find the perimeter of the square whose diagonal is measuring 4 cm.
Solution:
In the above square we have two right triangles. Those are ⊿ ACB and ⊿ ADC. In ⊿ ACB right angled at B. The side which is opposite to this angle is called hypotenuse side. We can find the other sides using Pythagorean theorem.
AC² = AB² + BC²
Since all sides are equal in square the sides AB and BC are equal in length.
Let AB = x so BC = x
AC² = x² + x² ==> 4² = 2x² ==> x² = 16/2 ==> x² = 8
x = √8
x = √2 x 2 x 2
x = 2 √2 cm
Hence, length of all sides = 2 √2 cm
Perimeter of square = 4a
= 4 (2 √2)
= 8 √2 cm
Hence, perimeter of new square is 8 √2 cm.
Question 10 :
The perimeter of two squares are 40 cm and 32 cm. Find the perimeter of third square whose area is equal to the difference of the area of two squares.
Solution :
Let "a" and "b" are side length of first and second squares respectively.
Perimeter of first square = 40 cm
4 a = 40 ==> a = 10
Perimeter of second square = 32 cm
4 b = 32 ==> b = 8
Area of third square = 10 ² - 8² ==> 36 cm²
Side length of third square = √36 ==> 6 cm
Perimeter of third square = 4 (6) = 24 cm
Hence, perimeter of third square is 24 cm
Question 11 :
The perimeter of five squares are 24 cm, 32 cm, 40 cm, 76 cm and 80 cm respectively. The perimeter of another square in area to the sum of the areas of those squares is:
Solution :
Area of new square = Sum of the areas of given squares
Perimeter of 1st square = 24 cm ==> 4a = 24 ==> a = 6
Side length of 1st square is 6 cm
Perimeter of 2nd square = 32 cm ==> 4b = 32 ==> b = 8
Side length of 2nd square is 8 cm
Perimeter of 3rd square = 40 cm ==> 4c = 40 ==> c = 10
Side length of 3rd square is 10 cm
Perimeter of 4th square = 76 cm ==> 4d = 76 ==> d = 19
Side length of 4th square is 19 cm
Perimeter of 5th square = 80 cm ==> 4e = 80 ==> e = 20
Side length of 5th square is 20 cm
The side lengths of given five squares are 6 cm, 8 cm, 10 cm, 18 cm and 20 cm respectively.
Sum of the areas of given squares = 6² + 8² + 10² + 19² + 20²
Area of new square = 961
Side length of new square = √961 = 31
Perimeter of new square = 4 (31) = 124 cm
Square Parallelogram |
Rectangle Rhombus |
Traingle Quadrilateral Sector
Hollow cylinder Sphere Area around circle Area of combined figures |
Trapezium Circle Semicircle Quadrant Cyclinder Cone Hemisphere Path ways |
WORD PROBLEMS
HCF and LCM word problems
Word problems on simple equations
Word problems on linear equations
Word problems on quadratic equations
Area and perimeter word problems
Word problems on direct variation and inverse variation
Word problems on comparing rates
Converting customary units word problems
Converting metric units word problems
Word problems on simple interest
Word problems on compound interest
Word problems on types of angles
Complementary and supplementary angles word problems
Markup and markdown word problems
Word problems on mixed fractrions
One step equation word problems
Linear inequalities word problems
Ratio and proportion word problems
Word problems on sets and venn diagrams
Pythagorean theorem word problems
Percent of a number word problems
Word problems on constant speed
Word problems on average speed
Word problems on sum of the angles of a triangle is 180 degree
OTHER TOPICS
Time, speed and distance shortcuts
Ratio and proportion shortcuts
Domain and range of rational functions
Domain and range of rational functions with holes
Graphing rational functions with holes
Converting repeating decimals in to fractions
Decimal representation of rational numbers
Finding square root using long division
L.C.M method to solve time and work problems
Translating the word problems in to algebraic expressions
Remainder when 2 power 256 is divided by 17
Remainder when 17 power 23 is divided by 16
Sum of all three digit numbers divisible by 6
Sum of all three digit numbers divisible by 7
Sum of all three digit numbers divisible by 8
Sum of all three digit numbers formed using 1, 3, 4
Sum of all three four digit numbers formed with non zero digits